Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $y = \dfrac{z + 10}{z^2 + 12z + 35} \times \dfrac{9z + 63}{-9z - 90} $
Explanation: First factor the quadratic. $y = \dfrac{z + 10}{(z + 7)(z + 5)} \times \dfrac{9z + 63}{-9z - 90} $ Then factor out any other terms. $y = \dfrac{z + 10}{(z + 7)(z + 5)} \times \dfrac{9(z + 7)}{-9(z + 10)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac{ (z + 10) \times 9(z + 7) } { (z + 7)(z + 5) \times -9(z + 10) } $ $y = \dfrac{ 9(z + 10)(z + 7)}{ -9(z + 7)(z + 5)(z + 10)} $ Notice that $(z + 10)$ and $(z + 7)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac{ 9(z + 10)\cancel{(z + 7)}}{ -9\cancel{(z + 7)}(z + 5)(z + 10)} $ We are dividing by $z + 7$ , so $z + 7 \neq 0$ Therefore, $z \neq -7$ $y = \dfrac{ 9\cancel{(z + 10)}\cancel{(z + 7)}}{ -9\cancel{(z + 7)}(z + 5)\cancel{(z + 10)}} $ We are dividing by $z + 10$ , so $z + 10 \neq 0$ Therefore, $z \neq -10$ $y = \dfrac{9}{-9(z + 5)} $ $y = \dfrac{-1}{z + 5} ; \space z \neq -7 ; \space z \neq -10 $